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4x^2+4x+1=9
We move all terms to the left:
4x^2+4x+1-(9)=0
We add all the numbers together, and all the variables
4x^2+4x-8=0
a = 4; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·4·(-8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*4}=\frac{8}{8} =1 $
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